cp's OEIS Frontend

n is necessarily composite.

From the Math. Rev.: if both q=7*2^{r-2}+2^s-1 and p=49*2^{2r-s-4}+7*2^{r-2}-5*2^{r-s-2}-1 are prime for 1 <= s < r-2, then n=2^r*3*p*q is a solution to the equation phi(n)+sigma(n)=3n . [R. K. Guy]

If 7*2^n-1 is prime then m = 2^(n+2)*3*(7*2^n-1) is in the sequence. Because phi(m) = 2^(n+2)*(7*2^n-2); sigma(m) = 7*2^(n+2)*(2^(n+3)-1) so phi(m)+sigma(m) = 2^(n+2)*((7*2^n-2)+(7*2^(n+3)-7)) = 2^(n+2)*(63*2^n-9) = 3*(2^(n+2)*3*(7*2^n-1)) = 3*m. Hence A112729 = 2^(A001771+2)*3*(7*2^A001771-1) is a subsequence of this sequence. - Farideh Firoozbakht, Dec 01 2005

If both numbers p=7*2^n+2^k-1 & q=49*2^(2n-k)+2^(n-k)*(7*2^k-5)-1 are prime and m=2^(n+2)*3*p*q then phi(m)+sigma(m)=3*m. Namely m is in the sequence. - Farideh Firoozbakht, Jan 11 2007

2*k = sigma(k) + phi(k) if and only if k is 1 or a prime.

If 7*2^j - 1 is prime then m = 2^(j+2)*3*(7*2^j - 1) is in the sequence. Because phi(m) = 2^(j+2)*(7*2^j - 2); sigma(m) = 7*2^(j+2)*(2^(j+3) - 1) so phi(m) + sigma(m) = 2^(j+2)*((7*2^j - 2) + (7*2^(j+3) - 7)) = 2^(j+2)* (63*2^(j+2) - 9) = 3*(2^(j+2)*3*(7*2^j - 1)) = 3*m, hence m is a term of A011251 and consequently m is a term of this sequence. A112729 gives such m's. - Farideh Firoozbakht, Dec 01 2005

Conjecture: For n > 1, a(n) is a Zumkeller number (A083207). Verified for all n in [2,63]. - Ivan N. Ianakiev, Jan 25 2023

The offset is 2, because for all numbers m, phi(m)+sigma(m) >= 2*m, so there is no number a(1) such that phi(a(1))+sigma(a(1))=1*a(1). - Farideh Firoozbakht, Jan 22 2008

a(5) >= 2*10^9. - Farideh Firoozbakht, Jan 22 2008

10^13 < a(5) <= 336280120525440. Charles R Greathouse IV showed that 6 divides a(5). 336280120525440 and 60493590969525342720 are the only m values I found such that phi(m) + sigma(m) = 5*m. - Donovan Johnson, Sep 11 2012